Kinematics

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Question 237 of 308

A body of mass 10kg10kg is projected at an angle of 45∘45^{\circ} with the horizontal. The trajectory of the body is observed to pass through a point (20,10)\left(20,10\right). If TT is the time of flight, then its momentum vector, at time t=T2t=\dfrac{T}{\sqrt{2}}, is _____ .
[Take g=10ms−2g=10m{s}^{-2}]

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Answer: (D) 1002i^+(1002−200)j^Ns100\sqrt{2}\hat{i}+\left(100\sqrt{2}-200\right)\hat{j}Ns

It is given that the projectile passes through (20,10)\left(20,10\right) and angle of projection is θ=45∘\theta =45^{\circ}. Using the trajectory equation,
y=xtan⁡θ−gx22u2cos⁡2θy=x\tan \theta -\dfrac{g{x}^{2}}{2{u}^{2}{\cos }^{2}\theta }
⇒10=20−(10)(100)u2\Rightarrow 10=20-\dfrac{\left(10\right)\left(100\right)}{{u}^{2}}
⇒u=20ms−1\Rightarrow u=20m{s}^{-1}Now the time of flight of projectile,
T=2usin⁡θgT=\dfrac{2u\sin \theta }{g}
⇒T=(2)(20)2(10)=22s⇒T2=2s\Rightarrow T=\dfrac{\left(2\right)\left(20\right)}{\sqrt{2}\left(10\right)}=2\sqrt{2}s \Rightarrow \dfrac{T}{\sqrt{2}}=2s
Now velocity of the projectile at any time tt can be written as,
v⃗=ucos⁡θi^+(usin⁡θ−gt)j^\vec{v}=u\cos \theta \hat{i}+\left(u\sin \theta -gt\right)\hat{j}
⇒v⃗=102i^+(102−10(2)]j^ms−1\Rightarrow \vec{v}=10\sqrt{2}\hat{i}+(10\sqrt{2}-10\left(2\right)]\hat{j}m{s}^{-1}
Therefore, momentum p⃗=Mv⃗=1002i^+(1002−200)j^Ns\vec{p}=M\vec{v}=100\sqrt{2}\hat{i}+\left(100\sqrt{2}-200\right)\hat{j}Ns

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